$\dfrac{ 10t + 2u }{ 9 } = \dfrac{ -9t + 6v }{ -4 }$ Solve for $t$.
Multiply both sides by the left denominator. $\dfrac{ 10t + 2u }{ {9} } = \dfrac{ -9t + 6v }{ -4 }$ ${9} \cdot \dfrac{ 10t + 2u }{ {9} } = {9} \cdot \dfrac{ -9t + 6v }{ -4 }$ $10t + 2u = {9} \cdot \dfrac { -9t + 6v }{ -4 }$ Multiply both sides by the right denominator. $10t + 2u = 9 \cdot \dfrac{ -9t + 6v }{ -{4} }$ $-{4} \cdot \left( 10t + 2u \right) = -{4} \cdot 9 \cdot \dfrac{ -9t + 6v }{ -{4} }$ $-{4} \cdot \left( 10t + 2u \right) = 9 \cdot \left( -9t + 6v \right)$ Distribute both sides $-{4} \cdot \left( 10t + 2u \right) = {9} \cdot \left( -9t + 6v \right)$ $-{40}t - {8}u = -{81}t + {54}v$ Combine $t$ terms on the left. $-{40t} - 8u = -{81t} + 54v$ ${41t} - 8u = 54v$ Move the $u$ term to the right. $41t - {8u} = 54v$ $41t = 54v + {8u}$ Isolate $t$ by dividing both sides by its coefficient. ${41}t = 54v + 8u$ $t = \dfrac{ 54v + 8u }{ {41} }$